Question

Two concentric circles are radii 5cm and 3cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.​

Answer 1

${\tt { {\underline{\underline{\huge{ AɴSᴡEʀ:}}}}}}$

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.

It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.

AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)

therefore, ΔODB is a right angled triangle

where, OD²+BD²=OB²

            (3)²+BD²=(5)²

             9+BD²=25

             BD²=25-9

             BD²=16

             BD=4cm

       Since AD=BD=4cm

       therefore, AB=AD+BD

                       AB=4+4

                       AB=8cm

Answer 2

Answer:

8 cm

Step-by-step explanation:

the radius of the smaller circle acts a a height for a right angle triangle and the radius of the larger circle acts as a hypotenuse if we join the 3 cm radius and 5 cm radius with the chord.

acc to pythagoras theorem, a^2 + b^2 = c^2

let x be the half of the chord we are finding (radius of 3 cm acts as a perpendicular to the chord)

so, x^2 + 3^2 = 5^2

x^2 = 25-9

x^2 = 16

x = + or - 4 but since a chord cannot be negative, x= 4 cm

now this is half of the chord we need, so, chord = 2 x 4 = 8 cm